Exercise 10.1: Pascal’s Triangle or Binomial Expansion
1. Find the expansion of $(1 + y)^5$ by the help of Pascal’s triangle or binomial theorem. With the help of the above expansions find (1) $(1 – y)^5$ and (2) $(1 + 2x)^5$.
Solution: Using the binomial expansion. we get,
$(1 + y)^5 = \binom{5}{0} + \binom{5}{1}y + \binom{5}{2}y^2 + \binom{5}{3}y^3 + \binom{5}{4}y^4 + \binom{5}{5}y^5$
$= 1 + 5y + 10y^2 + 10y^3 + 5y^4 + y^5$
(1) $(1 – y)^5$
Solution: Substituting $-y$ in place of $y$ in the above expansion. we get,
$(1 – y)^5 = 1 + 5(-y) + 10(-y)^2 + 10(-y)^3 + 5(-y)^4 + (-y)^5$
$= 1 – 5y + 10y^2 – 10y^3 + 5y^4 – y^5$
(2) $(1 + 2x)^5$
Solution: Substituting $2x$ in place of $y$ in the above expansion. we get,
$(1 + 2x)^5 = 1 + 5(2x) + 10(2x)^2 + 10(2x)^3 + 5(2x)^4 + (2x)^5$
$= 1 + 10x + 10(4x^2) + 10(8x^3) + 5(16x^4) + 32x^5$
$= 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5$
2. According to the ascending powers of $x$, expand the following up to first four terms (1) $(1 + 4x)^6$ and (2) $(1 – 3x)^7$.
Solution:
(1) First four terms of $(1 + 4x)^6$:
$(1 + 4x)^6 = \binom{6}{0} + \binom{6}{1}(4x) + \binom{6}{2}(4x)^2 + \binom{6}{3}(4x)^3 + \dots$
$= 1 + 6(4x) + \frac{6 \times 5}{2 \times 1}(16x^2) + \frac{6 \times 5 \times 4}{3 \times 2 \times 1}(64x^3) + \dots$
$= 1 + 24x + 15(16x^2) + 20(64x^3) + \dots$
$= 1 + 24x + 240x^2 + 1280x^3 + \dots$
(2) First four terms of $(1 – 3x)^7$:
$(1 – 3x)^7 = \binom{7}{0} + \binom{7}{1}(-3x) + \binom{7}{2}(-3x)^2 + \binom{7}{3}(-3x)^3 + \dots$
$= 1 + 7(-3x) + \frac{7 \times 6}{2 \times 1}(9x^2) + \frac{7 \times 6 \times 5}{3 \times 2 \times 1}(-27x^3) + \dots$
$= 1 – 21x + 21(9x^2) + 35(-27x^3) + \dots$
$= 1 – 21x + 189x^2 – 945x^3 + \dots$
3. Expand $(1 + x^2)^8$ first four terms. Find the value of $(1.01)^8$ by using the result.
Solution:
First four terms of $(1 + x^2)^8$:
$(1 + x^2)^8 = \binom{8}{0} + \binom{8}{1}(x^2) + \binom{8}{2}(x^2)^2 + \binom{8}{3}(x^2)^3 + \dots$
$= 1 + 8x^2 + \frac{8 \times 7}{2 \times 1}x^4 + \frac{8 \times 7 \times 6}{3 \times 2 \times 1}x^6 + \dots$
$= 1 + 8x^2 + 28x^4 + 56x^6 + \dots$
Now, comparing $(1 + x^2)^8 = (1.01)^8 = (1 + 0.01)^8$, we get:
$x^2 = 0.01 \implies x = 0.1$
Therefore,
$(1.01)^8 \approx 1 + 8(0.01) + 28(0.01)^2 + 56(0.01)^3$
$= 1 + 0.08 + 28(0.0001) + 56(0.000001)$
$= 1 + 0.08 + 0.0028 + 0.000056$
$= 1.082856$
Answer: Expansion: $1 + 8x^2 + 28x^4 + 56x^6 + \dots$ and Value: $1.082856$ (approx.).
4. According to the ascending powers of $x$, expand the following binomials upto first three terms
Solution:
(1) First 3 terms of $(1 – 2x)^5$:
$(1 – 2x)^5 = \binom{5}{0} + \binom{5}{1}(-2x) + \binom{5}{2}(-2x)^2 + \dots$
$= 1 + 5(-2x) + 10(4x^2) + \dots$
$= 1 – 10x + 40x^2 + \dots$
(2) First 3 terms of $(1 + 3x)^9$:
$(1 + 3x)^9 = \binom{9}{0} + \binom{9}{1}(3x) + \binom{9}{2}(3x)^2 + \dots$
$= 1 + 9(3x) + \frac{9 \times 8}{2 \times 1}(9x^2) + \dots$
$= 1 + 27x + 36(9x^2) + \dots$
$= 1 + 27x + 324x^2 + \dots$
5. Find the following expansions upto first four terms.
Solution:
(1) $(1 – 2x^2)^7$
$(1 – 2x^2)^7 = \binom{7}{0} + \binom{7}{1}(-2x^2) + \binom{7}{2}(-2x^2)^2 + \binom{7}{3}(-2x^2)^3 + \dots$
$= 1 + 7(-2x^2) + 21(4x^4) + 35(-8x^6) + \dots$
$= 1 – 14x^2 + 84x^4 – 280x^6 + \dots$
(2) $\left(1 + \frac{2}{x}\right)^4$
$\left(1 + \frac{2}{x}\right)^4 = \binom{4}{0} + \binom{4}{1}\left(\frac{2}{x}\right) + \binom{4}{2}\left(\frac{2}{x}\right)^2 + \binom{4}{3}\left(\frac{2}{x}\right)^3 + \dots$
$= 1 + 4\left(\frac{2}{x}\right) + 6\left(\frac{4}{x^2}\right) + 4\left(\frac{8}{x^3}\right) + \dots$
$= 1 + \frac{8}{x} + \frac{24}{x^2} + \frac{32}{x^3} + \dots$
(3) $\left(1 – \frac{1}{2x}\right)^7$
$\left(1 – \frac{1}{2x}\right)^7 = \binom{7}{0} + \binom{7}{1}\left(-\frac{1}{2x}\right) + \binom{7}{2}\left(-\frac{1}{2x}\right)^2 + \binom{7}{3}\left(-\frac{1}{2x}\right)^3 + \dots$
$= 1 + 7\left(-\frac{1}{2x}\right) + 21\left(\frac{1}{4x^2}\right) + 35\left(-\frac{1}{8x^3}\right) + \dots$
$= 1 – \frac{7}{2x} + \frac{21}{4x^2} – \frac{35}{8x^3} + \dots$
6. Expand (1) $(1 – x)^6$ and (2) $(1 + 2x)^6$ up to $x^3$ .
Solution:
(1) $(1 – x)^6$ (up to $x^3$):
$(1 – x)^6 = \binom{6}{0} + \binom{6}{1}(-x) + \binom{6}{2}(-x)^2 + \binom{6}{3}(-x)^3 + \dots$
$= 1 + 6(-x) + 15(x^2) + 20(-x^3) + \dots$
$= 1 – 6x + 15x^2 – 20x^3 + \dots$
(2) $(1 + 2x)^6$ (up to $x^3$):
$(1 + 2x)^6 = \binom{6}{0} + \binom{6}{1}(2x) + \binom{6}{2}(2x)^2 + \binom{6}{3}(2x)^3 + \dots$
$= 1 + 6(2x) + 15(4x^2) + 20(8x^3) + \dots$
$= 1 + 12x + 60x^2 + 160x^3 + \dots$